1 Lesson 19: Process Characteristics- 1 st Order Lag & Dead-Time Processes ET 438a Automatic Control Systems Technology 2 Learning Objectives After this series of presentations you will be able to: Describe typical 1 st order lag and dead-time process models found in control systems. Write mathematical formulas for 1 st order and deadtime process models Compute the parameters of process models. Identify the Bode plots of typical process models. Identify the time response of typical process models. 1
3 First Order Lag Process Characteristics: 1) Single storage element 2) Input produces an output related to amount of storage 3) Another name: self-regulating process Examples: Series R-C circuit Series R-L circuit Self-regulating tank (valve on output) Tank heating 4 First Order Lag Process Mathematical Descriptions Time domain equation: Transfer function: Equation Constants: dy t y G x dx Y s G X s 1 ts Dependent on System t determines time required for system to reach final value after step input 1t= 63.2% Final value 5t=99.3% Final value Where: G = steady-state gain of the system t = time, in seconds y = output of process (units or %FS) x = input of process (units or %FS) t = time constant of the system in seconds 2
5 First Order Lag Process Self-Regulating Tank Process Input Process input: q in = input flow rate (%FS in ) FS in = input range (m 3 /s) Process output: h = tank level, (% FS out ) FS out = output range (m) Process Parameters R L = flow resistance C L is the tank capacitance Process Output Equation Constants t R L C L R L FS G g FS in iout Where: = liquid density (kg/m 3 ) g = acceleration due to gravity 9.81 m/s 2 6 First Order Lag Process Self-Regulating Tank Example 19-1: Oil tank similar to previous figure has a diameter of 1.25 m and a height of 2.8 m. The outlet pipe is a smooth tube with a length of 5 m and diameter of 2.85 cm. Oil temperature 15 degrees C. The full scale flow rate is 24 L/min and full scale height is 2.8 m. Determine: a) tank capacitance, C L b) pipe resistance, R L c) process time constant, t d) process gain, G e) time-domain equation f) transfer function. 3
7 Example 19-1 Solution (1) a) Find the tank capacitance From Appendix A 8 Example 19-1 Solution (2) b) Find the flow resistance. First compute the Reynolds number to determine flow type. Use maximum flow to determine it. Convert to m 3 /s Need pipe diameter for its area calculation 4
9 Example 19-1 Solution (3) Now compute the Reynolds number Laminar Flow R<2000 10 Example 19-1 Solution (4) For Laminar flow Now compute the tank time constant Tank level reduced to 63.2% of initial value after 117.3 minutes with q in =0. 99.2% empty after 5t. 5
11 Example 19-1 Solution (5) d) Compute process gain, G Ans 12 Example 19-1 Solution (6) e) Find the differential equation for this process Ans f) Find the transfer function for this process Ans 6
Amplitude 11/13/2015 13 Step Response and Bode Plots of The First- Order Lag Process MatLAB Code % close all previous figures and clear all variables close all; clear all; % input the integral time constant Tl=input('enter the process time constant: '); G=input('enter the gain of the process: '); % construct and display the system sys=tf(g,[tl 1]); sys % plot the frequency response bode(sys); % construct a new figure and plot the time response figure; % define a range of time t=(0:500:5*tl); % use it to generate a step response plot step(sys,t); 14 Step Response of First-Order Lag Example 19-1 0.9 Process Step Response 0.8 0.7 0.6 0.5 0.4 0.3 0.632(0.818)=0.52 0.2 0.1 0 0 0.5 1 1.5 2 2.5 3 3.5 Time (seconds) x 10 4 t=7040 s 7
Magnitude (db) Phase (deg) 11/13/2015 15 Frequency Response of First-Order Lag Processes 0-10 -20-30 -40 0 Bode Diagram System: sys Frequency (rad/s): 0.000146 Magnitude (db): -4.87 Similar to low pass filter plots. Note the cutoff frequency, f c is when gain is -3 db from passband f=f c f c =1/t = 1.42 10-4 rad/sec -45-90 10-6 10-5 10-4 10-3 10-2 Frequency (rad/s) 16 First-Order Lag Process: Thermal Example Example 19-2: Temperature of oil bath q L depends on the steam temperature q J and the thermal resistance and capacitance of the system. The equation below is the general model. R T dql CT q dt L q Assume constant steam flow and temperature. Theo oil-filled tank is is 1.2 m tall with a 1 m diameter. The inside film coefficient is 62 W/m 2 -K and the outside film coefficient is 310 W/m 2 -K. The tank is made of steel with a wall thickness of 1.2 cm. Find a) thermal resistance b) thermal capacitance, c) thermal time constant d) differential equation model, e) transfer function model. J 8
17 Example 19-2 Solution (1) a) Compute the thermal resistance Unit thermal resistance 18 Example 19-2 Solution (2) Find the surface area of the tank. Assume a circular tank. Define: A 1 = area of tank bottom A 2 = area of tank sides Area of sides is area of a cylinder 9
19 Example 19-2 Solution (3) Compute total thermal resistance Ans b) Thermal capacitance S h found in Appendix A 20 Example 19-2 Solution (4) Ans Use the values of R T and C T to find time constant Step change in input will take 5t to reach final value. 5t 649 minutes (10.82 hours) 10
21 Example 19-2 Solution (5) d) Find the time function G=1 in this case so: Ans e) Find the transfer function Ans 22 Dead-Time Process Characteristic: Energy or mass transported over a distance Common in process industries (Chemicals Refining etc) Time domain equation: Transfer function: i f (t) f (t t ) t o d Fo (s) e F (s) D v i td s d Where: f o (t) = output function f i (t) = input function v = velocity of response travel (m/sec) D = distance from input to output t d = dead-time lag (sec or minutes) F o (s) = Laplace transform of output F i (s)= Laplace transform of input 11
23 Dead-Time Process Example: 19-3: Determine the dead-time lag and the process transfer function if the salt-water solution travels at 0.85 m/sec and the distance to the bend is 15 m. Plot the time and frequency response of this system to a step-change in inlet concentration. 24 Example 19-3 Solution (1) Define parameters Compute time delay Time function: Transfer function: t d v 0.85 m/sec D 15 m D v 15 m 0.85 m/sec c (t) c (t t o c (t) c (t 17.65) o Co(s) e C (s) i i i d ) 17.65s C(s) 17.65 sec C(s) is the function describing salt concentration Now plot the time and frequency responses 12
Phase Shift (Degrees) Gain (db) Concentration Level (0-1) Gain (db) 11/13/2015 25 Dead-Time Process Time Plot Dead-Time Process Time Response 1 c i ( t) c o ( t) 0.5 t d =17.65 sec 0 0 5 10 15 20 t Time (Seconds) Input Concentration Output Concentration 26 Dead-Time Frequency Bode Plot 0 Dead-Time Phase Plot 1 Dead-Time Gain Ploft 50 100 0.5 150 0 200 250 0.5 300 1 10 4 1 10 3 0.01 0.1 1 Frequency (rad/sec) 1 increases. 0.5 Becomes very large 0 Dead-Time Gain Ploft Phase increases as frequency for high frequencies. 1 1 10 4 1 10 3 0.01 0.1 1 Frequency (rad/sec) Gain is constant over all frequencies (0 db G=1) 13
27 ET 438a Automatic Control Systems Technology 14